Is discrete valuation field a fraction field of some Dedekind domain?

Question

Is discrete valuation field a fraction field of some Dedekind domain ?

Let $K$ be a discrete valuation field, does there exist some Dedekind domain $R$ such that $\operatorname{Frac}(R)$＝$K$ ？

If there are some counterexamples, it's also appreciated.

Thank you in advance.

Answer 1

Yes. For any valuation $v$ on the field $K$, with values in a totally ordered group $(G,+)$, the set $$R=\{x\in K\mid v(x)\ge 0 \}$$ is a local subring of $K$, with maximal ideal $\mathfrak m=\{x\in K\mid v(x) >0 \}$, and for any $x\in K\smallsetminus A$, $x^{-1}\in K$.